Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the value of n? Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the smallest value of n?
?What are the two numbers removed?
Now many answers are given but they are given in a hit and trial method.
Is there any other method to solve this problem?
 A: $\sum _{k=1}^n k= \dfrac{1}{2} n (n+1)$
Subtract $x+(x+1)$. The sum is then $\dfrac{1}{2} n (n+1) -2x-1$
The mean is then $\dfrac{\frac{1}{2} n (n+1) -2x-1}{n-2}=\dfrac{105}{4}$
$x=\dfrac{1}{8} \left(2 n^2-103 n+206\right)$ and $0<x<n$
$x>0 \to 2 n^2-103 n+206>0$ which means $n\ge 50$
And $x<n \to \dfrac{1}{8} \left(2 n^2-103 n+206\right)<n $ which means $2\le n\le 53$
so we have $50\le n\le 53$ but the only value that gives an integer $x$ is $n=50$
Finally the solution is $x=7$ when $n=50$
A: Since the given arithmetic mean in the title is $261/4$, we know that $(n-2)\times 261/4$ must be an integer, so $n\equiv 2 \bmod 4$.
The sum of the full set of integers from $1$ to $n$ is $n(n+1)/2$, with average $\color{blue}{(n+1)/2}$. Removing two numbers from that set cannot change the average by more than $1$, since removing the largest two makes it $(n-1)/2 = \color{blue}{(n+1)/2}-1$ and removing the two smallest makes it $(n-1)/2 + 2= \color{blue}{(n+1)/2}+1$.
So we know that $261/4 -1 \le \color{blue}{(n+1)/2} \le 261/4+1$ so $128.5\le n\le 132.5$. Now since only $130\equiv 2 \bmod 4$ in this range, we can be confident that  $n=130$ (assuming a well-posed question).
